 Good morning friends. I am Purva and today we will discuss the following question. Find the Cartesian equation of the line which passes through the point minus 2, 4, minus 5 and parallel to the line given by x plus 3 upon 3 is equal to y minus 4 upon 5 is equal to z plus 8 upon 6. Let vector A be the position vector of the given point. Let L be the line which passes through a point A and is parallel to vector B. Let vector R be the position vector of an arbitrary point P on the line. Then the vector equation of the line is given by vector R is equal to vector A plus lambda into vector B where lambda is the parameter. Here we have vector A is equal to x1 i cap plus y1 j cap plus z1 k cap and vector B is equal to a i cap plus b j cap plus c k cap. Here we have A, B and C are the direction ratios. Also we have vector R is equal to x i cap plus y j cap plus z k cap. Now putting the values of vector A, vector B and vector R in this equation 1 and eliminating lambda, we get the Cartesian form as x minus x1 upon A is equal to y minus y1 upon b is equal to z minus z1 upon c. So this is the key idea behind our question. Let us begin with the solution now. Now we have given that the line L is passing through the point minus 2 4 minus 5. So we get the position vector that is vector A is equal to minus 2 i cap plus 4 j cap minus 5 k cap. Now this is of the form vector A is equal to x1 i cap plus 5 1 j cap plus z1 k cap. So here we get x1 is equal to minus 2, y1 is equal to 4 and z1 is equal to minus 5. We mark this as 1. Now given Cartesian form of the line which is parallel to this line L is given by x plus 3 upon 3 is equal to y minus 4 upon 5 is equal to z plus 8 upon 6. Now this is of the form x minus x1 upon A is equal to y minus y1 upon b is equal to z minus z1 upon c where A, B and C are the direction ratios. That is coefficients of i cap, j cap and k cap in vector B which is parallel to the line L whose equation we have to find out. So on comparing them we can clearly see that A is equal to 3, here B is equal to 5 and C is equal to 6. So we have here A is equal to 3, B is equal to 5 and C is equal to 6. We mark this as 2. Now the equation of the line is given by x minus x1 upon A is equal to y minus y1 upon b is equal to z minus z1 upon c. Now putting the values of x1, y1, z1 and A, B, C from 1 and 2 in this equation we get x plus 2 upon 3 is equal to y minus 4 upon 5 is equal to z plus 5 upon 6. Thus we have got our answer as x plus 2 upon 3 is equal to y minus 4 upon 5 is equal to z plus 5 upon 6. Hope you have understood the solution. Bye and take care.